Question

Find the extreme values of the function F(x, y) = 3x2 + 5y? on the circle EXAMPLE 2 x2 + y2 = 1. SOLUTION We are asked for extreme values of f subject to the constraint 9(x, y) = x2 + y2 = 1. Using Lagrange multipliers, we solve the equations VF Ug and 9(x, y) = 1, which can be written as fx = 1gx fylgy (x,y) = 1 or as = 2x1 = 2ya x2 + y2 = 1. From (1) we have x = or i = 3. If x= then (3) gives y = $1. If I = 3, then y= from (2), so then (3) gives x = 1. Therefore F has possible extreme values at the points (0, 1), (0, -1), (1,0), and (-1,0). Evaluating fat these four points, we find that f(0, 1) = f(0, -1) = 5 F(1,0) = f(-1,0) = 3. Therefore the maximum value of fon the circle x2 + y2 = 1 is fo, +1) = and the minimum value is f(t1,0) = - Checking with the figure, we see that these values look reasonable.

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Answer 1

EXAMPLE 2 x2 + y2 = 1. Find the extreme values of the function F(x, y) = 3x2 + 5y2 on the circle of(x, y) => 286,7) > <6xe, log) = x <24,4) when x= 3, then from (2), ny = 2y (3) >> y=0 (1) (2) SOLUTION We are asked for extreme values of f subject to the constraint g(x, y) = x2 + y2 = 1. Using Lagrange multipliers, we solve the equations Of = 10g and 9(x, y) = 1, which can be written as fx = 1gx fy = igy g(x,y) = 1 or as 67€ roy x2 + y2 = 1. From (1) we have x = or 1 = 3. If x= then (3) gives y = 11. If i = 3, then y = from (2), so then (3) gives x = 41. Therefore f has possible extreme values at the points (0, 1), (0, -1), (1, 0), and (-1,0). Evaluating fat these four points, we find that f(0, 1) = 5 f(0, -1) = 5 F(1,0) = 1 f(-1,0) = 3. Therefore the maximum value of fon the circle x2 + y2 = 1 is f(0, 11) - 5 and the minimum value is f(+1,0) = 7 . Checking with the figure, we see that these values look reasonable. = 2x = 2yd f(0, 1) = 0 +5(12) = 5 +0,0) = 3(14) +0 = 3 fram = 5 st (0,11) f min = 3 at (£1,0)